POJ 3680 Intervals

Intervals

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 7424 Accepted: 3100

Description

You are given N weighted open intervals. The ith interval covers (a_i,b_i) and weighs w_i. Your task is to pick some of the intervals to maximize the total weights under the limit that no point in the real axis is covered more than k times.

Input

The first line of input is the number of test case.
The first line of each test case contains two integers, N and K (1 ≤K ≤ N ≤ 200).
The next N line each contain three integers a_i,b_i, w_i(1 ≤a_i < b_i ≤ 100,000, 1 ≤w_i ≤ 100,000) describing the intervals.
There is a blank line before each test case.

Output

For each test case output the maximum total weights in a separate line.

Sample Input

43 11 2 22 3 43 4 83 11 3 22 3 43 4 83 11 100000 1000001 2 3100 200 3003 21 100000 1000001 150 301100 200 300

Sample Output

1412100000100301

Source

【题目分析】

    一看到巨大的ai，bi和超级小的n，k就知道有必要进行离散化（就是因为离散化出错，RE了好多遍，有可能并没有重复，需要特判一下）。然后就不会做了，关键是每一个点覆盖不能超过k次，用网络流的方法确实是很巧妙的，加入（S，1，k，0）（T-1 ,T ,k ,0 ）（ai，bi，1,-wi）(i,i+1,k,0)如果一个区间被选择，那么这个区间可能通过的流量就减去了1（S点出来的流量为k，那么最大流就是k）然后就轻而易举的解决了。

【代码】

#include <cstdio>#include <cstring>#include <string>#include <queue>#include <algorithm>using namespace std;const int inf =0x3f3f3f3f;const int maxn=10010;const int maxm=10010;#define M(a) memset(a,-1,sizeof a)#define M0(a) memset(a,0,sizeof a)#define MB(a) memset(a,0x3f,sizeof a)int n,m,S,T,ans,en=0;int v[maxn],u[maxn],cost[maxn],f[maxn],h[maxn],ne[maxn];bool vis[maxm];//u->vint l[maxn],r[maxn],w[maxn],ss[maxn];int dis[maxm],with[maxm],minn[maxm];inline void add(int a,int b,int r,int c){u[en]=a;v[en]=b;ne[en]=h[a];f[en]=r;cost[en]=c; h[a]=en++;u[en]=b;v[en]=a;ne[en]=h[b];f[en]=0;cost[en]=-c;h[b]=en++;}inline bool tell(){queue<int>q;M0(vis);MB(dis);MB(minn);M0(with);q.push(S);vis[S]=true;dis[S]=0;while (!q.empty()){int x=q.front();q.pop();vis[x]=false;for (int i=h[x];i!=-1;i=ne[i]){int y=v[i];if (dis[y]>dis[x]+cost[i]&&f[i]>0){dis[y]=dis[x]+cost[i];minn[y]=min(minn[x],f[i]);with[y]=i;if (!vis[y]) vis[y]=1,q.push(y);}}}if (dis[T]==0x3f3f3f3f) return false;return true;}inline int zeng(){for (int i=T;i!=S;i=v[with[i]^1]){f[with[i]]-=minn[T];f[with[i]^1]+=minn[T];}return minn[T]*dis[T];}inline void solve(){M(v);M(u);M(h);M(ne);int n,k;scanf("%d%d",&n,&k);S=0;en=0;ans=0;for (int i=1;i<=n;++i){scanf("%d%d%d",&l[i],&r[i],&w[i]);ss[i*2-1]=l[i];ss[i*2]=r[i];}sort(ss+1,ss+2*n+1);for (int i=2*n;i>1;--i) if (ss[i]==ss[i-1]) ss[i]=inf;sort(ss+1,ss+2*n+1);int flag=0;for (int i=1;i<=2*n;i++){if (ss[i]==inf) {T=i+1;flag=1;break;}}if (!flag) T=2*n+1;for (int i=1;i<T;++i)add(i,i+1,k,0);for (int i=1;i<=n;++i){for (int j=1;j<T;++j) if (l[i]==ss[j]) {l[i]=j;break;}for (int j=1;j<T;++j) if (r[i]==ss[j]) {r[i]=j;break;}}for (int i=1;i<=n;++i){add(l[i],r[i],1,-w[i]);}add(S,1,k,0);add(T-1,T,k,0);while (tell()) ans-=zeng();printf("%d\n",ans);}int main(){int tt=0;scanf("%d",&tt);for (int z=1;z<=tt;++z) solve();}